If x, y and z are positive numbers, is \(\frac{x}{y} > \frac{x + z}{y + z}\) ?


(1) x > z

(2) x > y
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From given eqn we can deduce the following:
xy+xz>xy+yz
z(x-y)>0--(A)

so using 1: x>z , we are not given any relation about y so in -sufficient

From 1: X>Y
for any value of X >Y value the equation ( A) would stand true so B is sufficient..

There are two methods to solve this question.

Method 1:

Since x, y, z are all positive numbers, we can easily carry their denominators over the inequality without the fear of messing with the inequality.

Hence, \(\frac{x}{y} > \frac{x + z}{y + z}\) can be written as x(y+z)>y(x+z)
This implies xy+xz>xy+yz

Again, xy is a positive number here, so we cancel it out from both sides, leaving us with xz>yz. Taking the expression to one side of inequality, we get xz-yz>0

Thus, either z>0 and x>y or z<0 and x<y.

Now, we are given that z>0 (its a positive number). Hence, for the expression to be definitely true, we must also get x>y. And statement 2 does that for us.

Hence statement 2 is sufficient


Method 2:

Now, if we add the same positive number z to x and y in the fraction x/y, \(\frac{x + z}{y + z}\) will come closer to 1.

This gives us two possibilities:

1) If x/y<1, then \(\frac{(x+z)}{(y+z)}\) is greater than x/y

and

2) If x/y>1, then \(\frac{(x+z)}{(y+z)}\) is smaller than x/y

In our case, 2) works and this is satisfied only by statement 2. Hence, sufficient

Answer is B

Note: For anyone looking for more explanation on concept I used in statement 2, there is an article I think written by Karishma somewhere on GMATclub explaining this. If interested, I can look it up.